Q34 / 60 · JEE Advanced 2009 (Paper 1) · Mathematics
Paragraph:
Let $A$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1 . Five of these entries are 1 and four of them are 0 .Question:
The number of matrices $A$ in $A$ for which the system of linear equations $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ has a unique solution, is
- Aless than 4
- Correct answerBatleast 4 but less than 7
- Catleast 7 but less than 10
- Dat least 10
Solution
Given, $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
For unique solution, $\operatorname{det}(A) \neq 0$
$$
\begin{gathered}
\text { Case I } \operatorname{det}(A)=\left|\begin{array}{lll}
1 & a & b \\
a & 1 & c \\
b & c & 1
\end{array}\right| \\
=1-a^2-b^2-c^2+2 a b c \neq 0
\end{gathered}
$$
Here $a, b, c$ is selected from $1,0,0$. (No case is possible)
Case II
(i) $\operatorname{det}(A)=\left|\begin{array}{lll}1 & a & b \\ a & 0 & c \\ b & c & 0\end{array}\right|=2 a b c-c^2 \neq 0$
Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible)
(ii) $\operatorname{det}(A)=\left|\begin{array}{lll}0 & a & b \\ a & 1 & c \\ b & c & 0\end{array}\right|=2 a b c-b^2 \neq 0$
Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible)
(iii) $\operatorname{det}(A)=\left|\begin{array}{lll}0 & a & b \\ a & 0 & c \\ b & c & 1\end{array}\right|=2 a b c-a^2 \neq 0$
Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible)
Hence, there are exactly 6 matrices for unique solution. Hence, option (b) is correct.
Hence, the correct answer is option: 'B'.